换行符？Python

Question

我正在寻找仅用一个字符表示“\n”的方法。我正在写一个程序，它使用字典来“加密”文本。因为每个字符在字典中都有表示，所以当我的程序到达字符串中的“\n”时遇到问题，但将其读取为“\''n”。有没有其他方法来表示换行符，即只有一个字符？ 这是我下面的代码，如果缩进有问题，很抱歉。我不完全理解如何在这个窗口中输入代码。：）

##################################
#This program will take an input and encrypt it or decrypt it
#A cipher is used to transform the text, which can either be
#input or from a text file.
#The cipher can be any letter a-z, as well as most commonly used special characters
#numbers and spaces are not allowed.
#For the text, a-z, most special characters, space, and new line may be used.
#no numbers can be encrypted.
##################################

#These three dictionaries are used during the encryption process.
keyDict={'a': 17,'b': 3,'c':16,'d':26,'e':6,'f':19,'g':10,'h':12,
         'i':22,'j':8,'k': 11,'l':2,'m':18,'n':9,'o':23,'p':7,
         'q':5,'r': 20,'s': 1,'t': 24,'u':13,'v':25,'w':21,'x':15,
         'y':4,'z': 14, ' ':42, '.':0,'!': 27, '@': 34, '#': 35, '%': 37,
         '$': 36, "'": 33,'&': 39, '*': 40, ',': 29, '.': 30, '~': 41, ';': 31,
         ':': 32, '?': 28, '^': 38}

refDict2={' ': 43,  'a': 1, 'c': 3, 'b': 2, 'e': 5, 'd': 4, 'g': 7, 'f': 6, 'i': 9,
         'h': 8, 'k': 11, 'j': 10, 'm': 13, 'l': 12, 'o': 15, 'n': 14, 'q': 17,'\n': 42,
         'p': 16, 's': 19, 'r': 18, 'u': 21, 't': 20, 'w': 23, 'v': 22, 'y': 25,
         'x': 24, 'z': 26, ' ':0, '!': 27, '@': 34, '#': 35, '%': 37, '$': 36, "'": 33,
          '&': 39, '*': 40, ',': 29, '.': 30, '~': 41, ';': 31, ':': 32, '?': 28, '^': 38}

refDict={0: ' ', 1: 'a', 2: 'b', 3: 'c', 4: 'd', 5: 'e', 6: 'f', 7: 'g', 8: 'h', 9: 'i',
         10: 'j', 11: 'k', 12: 'l', 13: 'm', 14: 'n', 15: 'o', 16: 'p', 17: 'q', 42:'\n',
         18: 'r', 19: 's', 20: 't', 21: 'u', 22: 'v', 23: 'w', 24: 'x', 25: 'y', 26: 'z', 43:' ',
         32: ':', 33: "'", 34: '@', 35: '#', 36: '$', 37: '%', 38: '^', 39: '&', 40: '*',
         41: '~', 27: '!', 28: '?', 29: ',', 30: '.', 31: ';'}


#switch1 reverses a list. It is its own inverse, so we don't need an unswitch function. 
def switch1(l):
    return l[::-1]



#switch2 takes a list as input and moves every fourth entry to the front
#so switch2([a,b,c,d,e,f,g]) returns ([a,e,b,c,d,f,g])
#unswitch2 undoes this, so unswitch2([a,e,c,d,f,g]) returns [a,b,c,d,e,f,g]

def switch2(l):
    List4 = []
    ListNot4 = []
    for i in range(0,len(l)):
        if i%4 == 0:
            List4.append(l[i])
        else:
            ListNot4.append(l[i])
    return List4+ListNot4

def unswitch2(l):
    num4 = len(l)/4 + 1
    fixedList = l[num4:]
    for i in range (0,num4):
        fixedList.insert(4*i,l[i])
    return fixedList



#switch3 takes a list as input and returns a list with the first half moved to the end.
#so switch3([a,b,c,d,e,f]) returns [d,e,f,a,b,c]
#for lists of odd length, switch3 puts the separation closer to the beginning of the list, so the
#middle entry becomes the first entry.
#For example, switch3([a,b,c,d,e,f,g]) returns [d,e,f,g,a,b,c]

def switch3(l):
    return l[len(l)/2:] + l[:len(l)/2]

def unswitch3(l):
    if len(l)%2==0:
        return switch3(l)
    else:
        return l[len(l)/2+1:] + l[:len(l)/2+1]




##################################
#This is the Crypt function.
##################################
def Crypt(text, cipher):
    counter=0
    text=text.lower()
    cipher=cipher.lower()
    keyValue=[]
    textValue=[]
    newValue=[]
    newString=''
    for letter in cipher:
        keyValue.append(keyDict[letter])

    for letter in text:
        textValue.append(refDict2[letter])

    for num in textValue:
        newValue.append(num+keyValue[counter%len(keyValue)])
        counter+=1

    newValue = switch1(newValue)
    newValue = switch2(newValue)
    newValue = switch3(newValue)

    for num in newValue:
        newString+=refDict[num%43]

    return newString

##################################
#This is the Decrypt function
##################################
def Decrypt(encryptedText, cipher):
    counter=0
    cipher=cipher.lower()
    keyValue=[]
    textValue=[]
    finalValue=[]
    finalString=''

    for letter in encryptedText:
        textValue.append(refDict2[letter])

    textValue = unswitch3(textValue)
    textValue = unswitch2(textValue)
    textValue = switch1(textValue)

    for letter in cipher:
        keyValue.append(keyDict[letter])

    for num in textValue:
        finalValue.append((num-keyValue[counter%len(keyValue)])%43)
        counter+=1


    for num in finalValue:
        finalString+=refDict[num]

    return finalString


##################################
#This is the user interface.
##################################

choice=raw_input('Would you like to: 1)Encrypt or 2)Decrypt?  Pick 1 or 2: ')

if choice=='1':
    textType=raw_input("Would you like to: 1)encrypt a text file or 2) input the text to be encrypted? Pick 1 or 2: ")

    if textType=='1':
        cryptName=raw_input( 'Please enter the name of the text file you would like to encrypt(eg. text.txt): ')
        newName=raw_input('Please name the file in which the encrypted text will be stored(eg. secret.txt):' )
        cipher=raw_input("Now enter your personal encryption key(eg. secret code):" )

        cryptFile=open(cryptName, 'r')
        newFile=open(newName, 'w')
        print >> newFile, Crypt(cryptFile.read(),cipher)
        cryptFile.close()
        newFile.close()
        print "Ok, all done!"
    elif textType=='2':
        inputText=raw_input('Ok, please input the text you would like to encrypt(eg. computers rock!): ')
        cipher=raw_input("Now enter your personal encryption key (eg. ultra secret code): ")
        if inputText=='': print 'Oops, no text was entered! Try again!'
        else:
            print Crypt(inputText, cipher)
    else:
        print 'Try again!'

elif choice=='2':
    textType=raw_input("Would you like to:1)decrypt a text file or 2) input the text to be decrypted? Pick 1 or 2: ")
    if textType=='1':
        decryptName=raw_input( 'Please enter the name of the text file you would like to decrypt(eg. text.txt): ')
        newName2=raw_input('Please name the file in which the decrypted text will be stored(eg. secret.txt):' )
        cipher=raw_input("Now enter the encryption key that was used to encrypt the file(eg. secret code):" )
        #Text decrypt
        decryptFile=open(decryptName, 'r')
        newFile2=open(newName2, 'w')
        print>> newFile2, Decrypt(decryptFile.read(),cipher)
        #other stuff
        #textTodecrypt=decryptFile.read()
        #newFile2.writelines(Decrypt(textTodecrypt, cipher))


        decryptFile.close()
        newFile2.close()
        print "Ok, all done!"

    elif textType=='2':
        inputText=raw_input('Ok, please input the text you would like to decrypt(eg. dig&ak:do): ')
        cipher=raw_input("Now enter the encryption key that was used to encrypt the text (eg. ultra secret code): ")
        if inputText=='': print 'Oops, no text was entered! Try again!'
        else:
            print Decrypt(inputText, cipher)

print "Have a nice day!"


#there is an issue with the newline character 

Answer 1

我猜您真正的问题不是以'''''''n'的形式读入，在内部，Python应该自动将\n转换成一个字符。

我的猜测是，真正的问题是您的换行符可能实际上是两个字符——回车符（'\r'）和换行符（'\n'）。请尝试除处理外处理，我想知道这是否不会消除您的问题。

Answer 2

我假设问题是，如果要解密的文本中存在以下内容：

KeyError: \

module body   in untitled at line 166
function Crypt    in untitled at line 94

基本上，raw_input()返回一个包含两个字符\和n的字符串，并且没有\的映射，因此出现错误。

最简单的解决方案就是用\n转义序列替换文本字符\n

def Crypt(text, cipher):
    text.replace(r"\n", "\n")
    ... 

原始字符串r"\n"创建一个字符串，该字符串包含文本字符\，后跟n（与执行"\\n"相同）。

在常规字符串"\n"中，它被视为新行的转义序列。所以上面的代码块用换行符替换文本中的\n。

您可能必须在keyDict映射中为“\n”定义映射。

另一种解决方案是使用text.split(r"\n")拆分文本，并分别处理每一行。或者像其他人建议的那样，使用^{}每个字符并处理数字，而不是制作自己的数字映射。

ord(c) Given a string of length one, return an integer representing the Unicode code point of the character when the argument is a unicode object, or the value of the byte when the argument is an 8-bit string. For example, ord('a') returns the integer 97, ord(u'\u2020') returns 8224. This is the inverse of chr() for 8-bit strings and of unichr() for unicode objects.

正如文档所解释的，^{}正好相反，它会将数字转换回ASCII或Unicode字符：

chr(i) Return a string of one character whose ASCII code is the integer i. For example, chr(97) returns the string 'a'. This is the inverse of ord(). The argument must be in the range [0..255], inclusive; ValueError will be raised if i is outside that range.

基本上你会。。

def Crypt(text, cipher):
    keyvalue = [ord(x) for x in cipher)
    textvalue = [ord(x) for x in text]

…而不是两个for循环（list-comprehension基本上与创建列表相同，循环遍历文本或密码中的每个字符，并每次追加到列表中）

Answer 3

“\n”是一个字符。它是一个新行转义字符，只是新行的表示。

请用易读的方式重新表述你的问题。

[编辑] 我想我知道你的问题是什么。我运行了你的程序，一切正常。您可能正试图从命令行将'\n'传递给您的程序。那不行！

你看，如果你给raw_input（）这个字符串：line1\nline2它将转义\n，并使其成为\\n这样：'line1\\nline2'

所以一个快速的破解方法是找到并用“\n”替换“\n”：

text.replace('\\n', '\n')

我不喜欢这样。但这对你有用。

更好的方法是阅读多行，比如：

input = raw_input('Please type in something: ')
lines = [input]
while input:
    input = raw_input()
    lines.append(input)

text = '\n'.join(lines)

print text